BZOJ 3198([Sdoi2013]spring-Hash判重+容斥原理)

3198: [Sdoi2013]spring

Time Limit: 40 Sec  Memory Limit: 256 MB
Submit: 104  Solved: 42
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Description

Input

Output

Sample Input

3 3
1 2 3 4 5 6
1 2 3 0 0 0
0 0 0 4 5 6

Sample Output

2

HINT

Source

Dragonite修正数据

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Hash卡时间乱做——

遇到这种题目Hash玩桶排是和谐的

只要不和我似的全没long long就行

容斥部分简述

对于 111000 （3个1）而言

任意的4个1 （eg:1111000) 都会有C(4,3)遍 (0111000,1011000,1101000,1110000)

#include<cstdio>#include<cstdlib>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#include<cmath>#include<cctype>#include<map>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define RepD(i,n) for(int i=n;i>=0;i--)#define MEM(a) memset(a,0,sizeof(a))#define MEMI(a) memset(a,127,sizeof(a))#define MEMi(a) memset(a,128,sizeof(a))#define INF (2139062143)#define F (300007)#define MAXN (100000+10)#define BASE (13131)#define MAXK (6+1)int n,k; long long a[MAXN][7];int bin[7]={0,1,2,4,8,16,32};long long h[F+10],hnum[F+10][7];int st[MAXN],size=0;bool equal(long long h1[7],long long h2[7],int p){For(j,6) if (p&bin[j]) if (h1[j]^h2[j]) return 0;return 1; }int hash(int i,int p){long long ans=0;For(j,6) if (p&bin[j]) ans=(ans*BASE+a[i][j])%F;while (h[ans]>0&&(!equal(hnum[ans],a[i],p))) ans=(ans+1)%F;memcpy(hnum[ans],a[i],sizeof(hnum[ans])); return ans;}long long ans[MAXN]={0};int bitcount(int p){int i=0;For(j,6) if (bin[j]&p) i++;return i;}long long calc(int p){long long ans=0;size=0;For(i,n){int t=hash(i,p);h[t]++;st[++size]=t;}For(i,size)if (h[st[i]]){ans+=h[st[i]]*(h[st[i]]-1)/2;h[st[i]]=0;memset(hnum[st[i]],0,sizeof(hnum[st[i]]) );} return ans;}long long C[MAXK*2][MAXK*2]={0};int main(){//freopen("bzoj3197.in","r",stdin);//freopen("bzoj3197.out","w",stdout);scanf("%d%d",&n,&k);For(i,n) For(j,6) scanf("%lld",&a[i][j]);For(i,64/*(1<<6)-1*/){long long delta=calc(i);if (delta) ans[bitcount(i)]+=delta;//,cout<<i<<' ';}C[0][0]=1;For(i,12){C[i][0]=1;C[i][1]=i;Fork(j,2,12) C[i][j]=C[i-1][j]+C[i-1][j-1];}//cout<<C[2][0]<<' '<<C[2][1]<<' '<<C[2][2]<<endl;//Rep(i,7) cout<<ans[i]<<endl;ForkD(i,k,6) {Fork(j,i+1,6) ans[i]-=C[j][j-i]*ans[j];}//Rep(i,7) cout<<ans[i]<<endl;cout<<ans[k];return 0;}